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Homework Help: Acid base reations

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data
    after titrating sodium hydroxide with hydrofluoric acid, a chemist determined that the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride. determine ph of the solution

    2. Relevant equations

    3. The attempt at a solution
    Ka of HF= 6.3X10^-4
    NaOH + HF (equlibrium arrows) NaF +H20
    I ? ? 0 0
    C -X -X +X +X
    E ?-X ?-X 0.020 0.020

    ka = product/reactants i got stuckhere because there no initial value, and i don't know why im trying to find initial when i need to find ph
  2. jcsd
  3. Jan 2, 2010 #2


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    Staff: Mentor

    You look for pH of 0.020M F- solution, everything else doesn't matter. This is a weak base.

  4. Jan 2, 2010 #3
    wait i don't get it ? -log 0.020M is 1.69 that is wrong b/c Strong base reacts with weak acid so ph should be aboe 7 b/c it will be a basic solution
  5. Jan 2, 2010 #4
    You don't need to know I, the initial, since we are given that the Equilibrium of NaF (C-X) is 0.020

    The dissociation equation, as you said is:

    NaF + H20 <==> NaOH + HF
    (B + H2O <==> HB + OH-)


    F- + H2O <=> HF + OH-

    I C ........N/A.............
    C C -X .......... +X .... +X
    E 0.020 .......... +X .... +X

    Since we have B + H2O <==> HB + OH-, the equation to use is then Kb=[HB][OH-]/[B-] so we need to get Kb from Ka so...

    Kb = Kw/Ka = [10^-14]/[6.3x10^-4] = 1.58 x10^-11​

    We now know:
    Kb=1.58 x10^-11

    All you have to do is plug these values into Kb=[HF][OH-]/[F-] and solve for X
    Remember... X=[OH-], so you need to get the pH from the pOH

    Relevant equations:
    B + H2O <==> HB + OH-
    Kb = Kw/Ka
    pOH= -log(OH-)
    pOH + pH=14
    Last edited: Jan 3, 2010
  6. Jan 3, 2010 #5
    thanks so much!! you just made my day.
  7. Jan 3, 2010 #6


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    Staff: Mentor

    No, we are given that initial F- is 0.020M, and equilibrium concentration of HF is not 0.020M but has to be calculated. And in fact it is a lot smaller than 0.020M.

    You will be badly surprsised if you assume x12179x approach.

  8. Jan 3, 2010 #7
    Hmm, I mistyped HF instead of NaF above (and also .2 instead of .02); Now corrected

    But I thought 0.02 was the equilibrium value of NaF since it said:

    "after titrating...the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride." ?
  9. Jan 3, 2010 #8
    thats kinda what i assumed to
  10. Jan 4, 2010 #9


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    Staff: Mentor

    Just because you have 0.020M solution of NaF doesn't mean 0.020M is an equilibrium concentration of F-. Quite the opposite - F- is a weak Broensted base, reacting with water, so its concentration after hydrolysis is smaller.

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